Set, Relations and FunctionsMTP June 2023 Series IIQuestion 1998 of 217
All Questions

If f(x)=x1+x2\displaystyle f(x)=\frac{x}{\sqrt{1+x^2}} and g(x)=x1−x2\displaystyle g(x)=\frac{x}{\sqrt{1-x^2}} Find fog?\displaystyle fog?

Options

Ax\displaystyle x
B1/x\displaystyle 1/x
Cx/1−x2\displaystyle x/\sqrt{1-x^2}
Dx/1−x2\displaystyle x/\sqrt{1-x^2}
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Correct Answer

✅ Option a — x\displaystyle x

All Options:

  • Ax\displaystyle x
  • B1/x\displaystyle 1/x
  • Cx/1−x2\displaystyle x/\sqrt{1-x^2}
  • Dx/1−x2\displaystyle x/\sqrt{1-x^2}

Detailed Solution & Explanation

To find the composite function (f∘g)(x)=f(g(x))\displaystyle (f \circ g)(x) = f(g(x)), we substitute g(x)\displaystyle g(x) into f(x)\displaystyle f(x):
Given:
f(x)=x1+x2f(x) = \frac{x}{\sqrt{1 + x^2}}
g(x)=x1−x2g(x) = \frac{x}{\sqrt{1 - x^2}}

Substitute g(x)\displaystyle g(x) in place of x\displaystyle x in f(x)\displaystyle f(x):
f(g(x))=g(x)1+(g(x))2=x1−x21+(x1−x2)2f(g(x)) = \frac{g(x)}{\sqrt{1 + (g(x))^2}} = \frac{\frac{x}{\sqrt{1 - x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 - x^2}}\right)^2}}

**Step 1: Simplify the term inside the radical in the denominator**
1+(x1−x2)2=1+x21−x2=(1−x2)+x21−x2=11−x21 + \left(\frac{x}{\sqrt{1 - x^2}}\right)^2 = 1 + \frac{x^2}{1 - x^2} = \frac{(1 - x^2) + x^2}{1 - x^2} = \frac{1}{1 - x^2}

**Step 2: Take the square root of the simplified expression**
11−x2=11−x2\sqrt{\frac{1}{1 - x^2}} = \frac{1}{\sqrt{1 - x^2}} (where ∣x∣<1\displaystyle |x| < 1)

**Step 3: Substitute back and simplify**
f(g(x))=x1−x211−x2=x1−x2⋅1−x21=xf(g(x)) = \frac{\frac{x}{\sqrt{1 - x^2}}}{\frac{1}{\sqrt{1 - x^2}}} = \frac{x}{\sqrt{1 - x^2}} \cdot \frac{\sqrt{1 - x^2}}{1} = x

Hence, **Option A** is the correct answer.

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